# Passing a 2 dimensional array as a function parameter

I need to write a function that recieves a two dimensional array as a parameter. The problem is that the size of both dimensions are unknown, and will change each time the function is called. I started out writing it like this...

main()
{
double lfArray[25][8];
int nSizeDim1 = 25;
int nSizeDim2 = 8;

someFunction(lfArray, nSizeDim1, nSizeDim2);
}

void someFunction(double** lfArray, int nSizeDim1, int nSizeDim2)
{
do something with the array;
}

Then I changed the calling function to look like...
someFunction((double**)lfArray);

Using the cast to the double** the program compiled, but the data in the two dimensional array was not accessible from the someFunction.

Any help would be greatly appreciated.
Thanks
###### Who is Participating?

x

Analyst ProgrammerCommented:
One way to solve your problem is to allocate memory to "lfArray" dynamically.

double **lfArray;
int nSizeDim1 = 25;
int nSizeDim2 = 8;

lfArray=new double*[25];
for (int i=0; i<25; i++)
lfArray[i]=new double[8];

Then you don't need casting when passing to function.
0

Commented:
hmm...  What about using std vector?  Good luck...
0

Commented:
Calling

pszResult = new DWORD*[10];
pszResult[0] = new DWORD;
etc

someFunction((DWORD*&)pszResult)

//////////////////////////////////

someFunction(DWORD* &pBufferResult, int nSizeDim1, int nSizeDim2)
{
DWORD **ppdwArray = (DWORD**)pBufferResult;

ppdwArray[index][index];

do something with the array;
}

GOOD LUCK
0

Commented:
void someFunction(double** lfArray, int nSizeDim1, int nSizeDim2)
{
for( int row=0; row< nSizeDim1; row++)
for( int col =0; col < nSizeDim2; col++ )
// element ( row, col ) = 0
lfArray[ row*nSizeDim2 + col] = 0
}
0

Commented:
Sorry, fucntion prototype should be

void someFunction(double* lfArray, int nSizeDim1, int nSizeDim2);

double* lfArray; //not double** lfArray

void someFunction(double* lfArray, int nSizeDim1, int nSizeDim2)
{
for( int row=0; row< nSizeDim1; row++)
for( int col =0; col < nSizeDim2; col++ )
// element ( row, col ) = 0
lfArray[ row*nSizeDim2 + col] = 0
}

Note :

Assume we have

double arr[SizeRow][SizeCol];

We wil have

&arr[row][col] = ((double *) arr) + row*SizeCol + col

0

Commented:
Here is a example code

#include <iostream>
#include <iomanip>

using namespace std;

void someFunction(double* lfArray, int nSizeRow, int nSizeCol)
{
double value = 0.;
for( int row =0; row < nSizeRow; row++)
for( int col =0; col < nSizeCol; col++)
{
lfArray[row*nSizeCol + col] = value;
value = value+1;
}

}

main()
{
double lfArray[3][8];
int nSizeDim1 = 3;
int nSizeDim2 = 8;

someFunction( (double *)lfArray, nSizeDim1, nSizeDim2);

for( int row =0; row < nSizeDim1; row++)
{
for( int col =0; col < nSizeDim2; col++)
cout << setw(3) << lfArray[row][col] << "  ";
cout << endl;
}

}

The output wil be

0    1    2    3    4    5    6    7
8    9   10   11   12   13   14   15
16   17   18   19   20   21   22   23
0

Author Commented:
Worked great.

I tried figuring out the vector approach, but couldnt find any info on two dimensional vectors. And I forgot to mention (and show in my example) that the two dimensional array would be full of data BEFORE passing to the function, so some of the other comments didnt apply.

Thanks to everyone for their assistance.
0

Commented:
>> I tried figuring out the vector approach,
I would strongly recommend a vector or other class for solving this problem.  The solutions suggested will certainly work, but are invitations for bugs.  If you never make mistakes, they are fine.  If you are human...I'd try something else.

You can easily create a 2D vector array by using a vector of vectors.  However, each of the vector's contained in the outer vector cna have different sizes.  i.e. the array may not be a "rectangle".  (This is true of the current solution too.)  However, sometimes this is actually disirable and often it is not a problem.

anyways you can do this like

void somefunction (const  vector< vector<int> > &V)
{
for (int y = 0; y < V.size(); ++x)
{
for (int x = 0; x < V[y].size(); ++x)
cout << V[y][x];
cout << endl;
}
};

Note that one common mistake is to not include a space between the >'s that apear in the nested tempalte declaration.   For example

const  vector< vector<int>> &V    // Error, needs a space.
const  vector< vector<int> > &V  // Okay.

the two >'s together  form a single >> operator, and according to the max-munch principle, must be parsed as a single operator, not two sperate ones.
0
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